中序遍历序列与前序遍历序列构造二叉树_头歌inpretotree(): 由前序遍历序列和中序遍历序列构造二叉树前序序列为pa[p1:p2-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_36429070/article/details/108918975

该博客介绍了如何利用后序遍历序列（postorder）重建二叉树，通过递归方法，结合前序遍历（preorder）和中序遍历（inorder）的关系，实现从后序序列重构树的完整过程。

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int sz = preorder.size();
        return build(preorder, 0, sz - 1, inorder, 0, sz - 1);
    }
    TreeNode* build(vector<int>& preorder, int prestart, int preend, vector<int>& inorder, int instart, int inend){
        //这一步很关键,因为前序序列是靠中序序列pivol两侧的长度来截取,在递归中迟早会使头超过尾
        if(prestart > preend)
            return NULL;
        int sz = preorder.size();
        int rootVal = preorder[prestart];
        int pivol;
        for(int i=0;i<sz;i++){
            if(inorder[i]==rootVal){
                pivol = i;
                break;
            }
        }
        int leftsz = pivol - instart;
        TreeNode* root = new TreeNode(rootVal);
        root->left = build(preorder, prestart + 1, prestart + leftsz, inorder, instart, pivol - 1);
        root->right = build(preorder, prestart + leftsz + 1, preend, inorder, pivol + 1, inend);
        return root;
    }
};

后序遍历序列的不同在于，取的是序列的最后一位。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        int sz = inorder.size();
        return build(inorder, 0, sz - 1, postorder, 0, sz - 1);
    }
    TreeNode* build(vector<int>& inorder, int inStart, int inEnd, vector<int>& postorder, int postStart, int postEnd){
        if(postStart > postEnd)
            return NULL;
        int sz = inorder.size();
        int rootVal = postorder[postEnd];
        int pivol;
        for(int i=0;i<sz;i++){
            if(inorder[i]==rootVal){
                pivol = i;
                break;
            }
        }
        int leftsz = pivol - inStart;
        TreeNode* root = new TreeNode(rootVal);
        root->left = build(inorder, inStart, pivol - 1, postorder, postStart, postStart + leftsz - 1);
        root->right = build(inorder, pivol + 1, inEnd, postorder, postStart + leftsz, postEnd - 1);
        return root;
    }
};